Using Hund's rule, how would you write the ground state electronic configurations for the following Ions: a) #O^+# b)#C^-# c)#F^+# d)#O^-2#?
1 Answer
#"O"^(+): 1s^2 2s^2 2p^3#
#"C"^(-): 1s^2 2s^2 2p^3#
#"F"^(+): 1s^2 2s^2 2p^4#
#"O"^(2-): 1s^2 2s^2 2p^6#
NEUTRAL ATOMS
Well, we first need the neutral configurations to start from...
...and we have them here:
#bb("C": 1s^2 2s^2 2p^2)" "" "" "" "" "bb("O": 1s^2 2s^2 2p^4)#
#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "" "ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))#
#underbrace(" "" "" "" "" "" "" "" ")" "" "underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "2p" "" "" "" "" "" "" "" "" "" "2p#
#ul(uarr darr)" "" "" "" "" "" "" "" "ul(uarr darr)" "#
#2s" "" "" "" "" "" "" "" "" "" "2s#
#" "#
#bb("F": 1s^2 2s^2 2p^5)#
#ul(uarr darr)" "ul(uarr darr)" "ul(uarr color(white)(darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "2p#
#ul(uarr darr)" "#
#2s#
And so, by conservation of charge,
#"O"^(+)# has one less electron than#"O"# .#"C"^(-)# has one more electron than#"C"# .#"F"^(+)# has one less electron than#"F"# .#"O"^(2-)# has two more electrons than#"O"# .
CONFIGURATION OF
The hardest one here to consider when adding electrons is
#bb("C"^(-): 1s^2 2s^2 2p^color(red)(3))#
#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(red)(uarr) color(white)(darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "2p#
#ul(uarr darr)" "#
#2s#
And we did not pair it up, since that would introduce unnecessary electron repulsion. Instead, we place it in the last empty
It is thus more stable to leave all electrons unpaired until we have already achieved all half-filled orbitals in the same sublevel. And this, we call Hund's Rule.
CONFIGURATION OF
The hardest one here to consider when removing electrons is
#bb("F"^(+): 1s^2 2s^2 2p^color(red)(4))#
#ul(uarr darr)" "ul(uarr cancel(color(red)(darr)))" "ul(uarr color(white)(darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "2p#
#ul(uarr darr)" "#
#2s#
(We could have, then, three degenerate configurations. What are the other two?)
THE REST?
From these, you should be able to handle the rest by adding or subtracting electrons from the above configurations already shown for the neutral atoms.
- The electron configuration of
#"O"^(+)# will be identical to#"C"^(-)# . You're already done... #"O"^(2-)# has everything paired, so just pair up your electrons without overfilling a given orbital. This will be the easiest one.
